EE 261 The Fourier Transform and its Applications Fall 2006 Midterm Exam Solutions

Transcription

1 EE 6 The Fourier Transform and its Applications Fall 006 Midterm Exam Solutions There are six questions for a total of 00 points. Please write your answers in the exam booklet provided, and make sure that your answers stand out. Don t forget to write your name on your exam book!

2 . (5 points) Let f(t) be a periodic signal of period and define the averaging operator depending on a parameter h>0by Thus A h f(x) is a new signal. A h f(x) = h x+h x h f(t) dt. (a) Show that A h f(x) is periodic of period as a function of x, i.e., A h f(x +)=A h f(x), Hint: A h f(x +)= x++h f(t) dt. h x+ h Now make a change of variable t = u +. (b) Find the Fourier series for A h f(x) in terms of the Fourier series for f(t). Solution: To show that A h f is periodic of period we have to work directly with the integral, A h f(x +)= h x++h x+ h f(t) dt. To bring the periodicity of f(t) into the picture we make a change of variable t = u +in the integrand. The limits of integration change according to t = x + h u = x h and t = x ++h u = x + h. We find A h f(x +)= x++h f(t) dt h x+ h = x+h f(u +)du h x h = x+h f(u) du (since f(t) is periodic of period ) h x h = A h f(x). Next write the Fourier series for f(t) as f(t) = c n e πnit. We apply A h to f(t) by integrating the series term by term: x+h A h f(x) = c n e πnit dt h x h and it s the integral we have to evaluate. For this, if n = 0 then h x+h x h dt = h ((x + h) (x h)) = h h =.

3 If n 0, h x+h x h Thus the Fourier series for A h f(x) is e πnit dt = [ ] t=x+h h πin eπnit t=x h = 4πinh (eπin(x+h) e πin(x h) ) = 4πinh eπinx (e πinh e πinh ) = e πinh e πinh e πinx πnh i sin πnh = πnh eπinx = sinc(nh)e πinx. A h f(x) = c n sinc(nh)e πinx. Incidentally, the resulting expression obtained by integrating the Fourier series term by term in this way also shows that A h f(x) is periodic of period. 3

4 . (5 points) Energy of a bandlimited signal The energy of a signal g(t) is the integral Suppose that g(t) is bandlimited with g(t) dt. Fg(s)=0, s. Express the energy of g(t), in terms of the sample values of g(t) at the integers, g(n), n = 0, ±, ±,... Hint: Use the Fourier series of the periodization of Fg(s) of period. Solution: Write the Fourier series of the periodized version of Fg(s) as The coefficients are given by c n = = / / = g( n) e πins Fg(s) ds c n e πins. e πins Fg(s) ds (since Fg(s) = 0 for s /) = F Fg( n) Thus the Fourier series is By Parseval s identity g( n)e πint and by Rayleigh s identity / / g(t) dt = Ff(s) ds = Ff(s) ds = / / Ff(s) ds, g( n) = g(n). This allows us to express the energy in terms of the samples: g(t) dt = g(n). 4

5 3. (5 points) Let f(t) be a signal. Its Fourier transform F (s) is plotted below. 3 F (s) 0 3 Sketch the graph of the Fourier transform of the signals: (a) f( x) (b) f(x) (c) e 4πix f(x) (d) (f f)(x) (e) πi f (x) Solutions: (a) One of the basic duality results tells us that F(f( x)) = F ( s), the reverse of the Fourier transform. The graph looks like this. F ( s)

6 (b) According to the stretch theorem F(f(x)) = (/)F (s/). The plot is: 3/ (/)F (s/) / (c) A phase change in time corresponds to a shift in frequency, F(e 4πix f(x)) = F (s ), in this case a shift to the right by. The plot is: 3 F (s ) (d) Under the Fourier transform convolution goes to multiplication, so The graph is: F(f f) =(Ff). 6

7 9 F (s) (e) The derivative theorem gives F ( ) πi f (x) = sf (s). The graph is: 9 6 sf (s)

8 4. (5 points) Let f(t) be a signal with Fourier transform F (s). Suppose we are given the following facts: (a) f(t) is real. (b) f(t) = 0 for t 0. (c) F (Re F (s)) = t e t, where Re F (s) denotes the real part of F (s). Find f(t). Hint: Recall that Re F (s) =(F (s)+f(s))/. Solution: From the hint, F (s)+f(s) Re F (s) = and since f(t) is real its Fourier transform has the symmetry F (s) =F ( s). Thus and we have Re F (s) = t e t = F (Re F (s)) F (s)+f ( s) = (F (F (s)) + F (F ( s)) = (f(t)+f( t)). Now, if t>0 then t <0 and te t = t e t = (f(t)+f( t)) = f(t). We can then say or simply f(t) = { te t, t > 0, 0, t 0 f(t)te t H(t), where H(t) is the unit step function. 8

9 5. (5 points) Stand a fixed distance from a wall, hold a laser horizontally, and point it at the wall at a random angle. The locations of the burn marks on the wall is a random variable X whose values are distributed according to p(x) = π( + x ). This is a Cauchy distribution and arises in many applications. (a) Find the distribution for the average of two (independent) series of such burn marks, i.e. (/)(X + X ), where X and X are distributed as above. For this you ll likely need the Fourier transform of a two-sided decaying exponential (see the formula sheet), and the fact that if p(x) is the distribution for a random variable X then (/a)x has distribution ap(ax). (b) Without any further work, what happens if we average N series of such burn marks? (c) People s intuition is often that if they perform the same experiment many times and average the results then they re getting close to the actual or ideal answer. What do your results above say about this intuition? Solutions: (a) The distribution for the sum of two independent random variables is the convolution of their distributions. In our case the sum we want is X + X and the distribution for each is p(x)= π( + (x) ). To find the convolution of this with itself we take the Fourier transform to convert to a product. For this we need the formula Fe a t a = a +4π s, where by duality we have ( ) a F a +4π s = e a t. To get this to match with our distribution p(x) we take a = π, π π +4π x = π( + (x) ). Then {( ) ( )} π π F π +4π x π +4π x =(e π s )(e π s )=e π s. The inverse Fourier transform of this is 4π 4π +4π x = π( + x ) = p(x). We re back to p(x)! 9

10 (b) Thus we see that the distribution of the average of two Cauchy random variables is the same Cauchy random variable. From this, the average of N Cauchy Random variables is the same as that of the first Cauchy random variable. (c) However many times we repeat the experiment, on averaging the results we ll still find the same distribution. There is no improvement. 0

11 6. (5 points) The three TA s are out having dinner one night and talk EE6 and sampling theory. Liz: So I want to sample this signal at 3 samples per second, but I only have samplers that operate at sample per second. I ought to be able to start them a third of a second apart and add the results together to make three samples a second. That will allow me to reconstruct a signal with bandwidth.5 Hz. Thomas: Well that sounds good, but if you take the Fourier Transform of the second samplers you will get three Shah functions, and if you add those up you will get the sum of impulse trains each spaced hertz apart. Won t you? And that s only going to allow you to reconstruct a signal with bandwidth 0.5 Hz. Paul: What is going on here? Who is right, and why is the other person wrong? And who is picking up the check? Solution Liz is correct, combining three shifted samples one third second apart will give an impulse train with impulses /3 seconds apart. III(t) + III(t 3 )+III(t 3 ) = III 3 (t). The Fourier transform of the right hand side is FIII /3 = 3III 3. Now here s what happens in Thomas s objection. The Fourier Transform of the sum of the shifted III s is III(s)+e πi 3 s III(s)+e πi 3 s III(s) = III(s)( + e πi 3 s + e πi 3 s ) Multiplying the terms in III by the factor out front gives e πis/3 If n is an integer not a multiple of 3 then = (e πis/3 ) 3 III(s) (using the sum of a geometric series) e πis/3 = e πis III(s) e πis/3 = e πis/3 sin πs sin(πs/3) III(s) sin πs sin πn δ(s n) =e πin/3 δ(s n). sin(πs/3) sin(πn/3) e πin/3 sin πn sin(πn/3) =0 so those terms are killed off. If n is a multiple of 3 then e πin/3 = and sin πn sin(πn/3) =3.

12 sin x (To see this last point, it s like lim =/(/3) = 3.) Thus we only get every third x 0 sin(x/3) impulse, multiplied by 3: e πis/3 sin πs sin(πs/3) III(s) = 3III 3(s) just as before.